Differential Equations - Method of Undetermined Coefficients

[MJay82]

Homework Statement

I've been having problems with a number of these things, here's the first one:
y'' -2y' -3y = -3te^-t

Homework Equations

I know that the general solution will be
y = y_h + y_p
where y_h is the general solution to the homogeneous equation, and y_p
is the particular solution of the non-homogeneous equation.

The Attempt at a Solution

I got the homogeneous solutions very easily, but I'm tricked by how to solve for y_p. I understand the principle of the solution if the right side of the equation was simply -3e^-t, but when the extra t is thrown in there, my understanding breaks down.

I tried y_p=t²e^-t, differentiated twice, and then input these expressions into the equation. I just realized that I was supposed to use an unknown coefficient (A we'll call it) with my guess at y_p. Since the terms on the right side are a product, will I just use A, or will I need some B as well? Thanks for any help.

 

[ehild]

You need to use unknown coefficients, in plural . Try the solution in the form y_p=(at²+bt+c)e^-t.

ehild

 

[HallsofIvy]

IF $e^{-t}$ were not already a solution to the associated homogeneous equation, since the right side is $te^{-t}$, you would try $y_p= (At+ B)e^{-t}$. Because $e^{-t}$ IS a solution to the associated homogeneous equation, you should try $y_p= (At^2+ Bt)e^{-t}$.

(You don't really need the "c" in ehild's suggestion. It wouldn't hurt, but you would find that c= 0.)

 

[MJay82]

Thanks y'all! I'll be trying a little later this afternoon.